Log in
HomeClass Notes1,200,000CA670,000

MCB 2050 Lecture Notes - Lecture 1: Base Pair, Adenosine Monophosphate, Gel Electrophoresis

68 views14 pages
bluehyena388
12 Jan 2018
School
University of Guelph
Department
Molecular and Cellular Biology
Course
MCB 2050
Professor
Dick Mosser

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

Dna molecules can be cut at specific sites with restriction enzymes and joined together using. Dna is identical but inverted in the complementary strand. Construction of recombinant dna molecules in vitro (in test-tube) Step 1: cut plasmid and dna you want to clone with restriction enzyme (ecori) Step 2: anneal and ligate complementary dna fragments. Dna molecules can be cut at specific sites with restriction enzymes and joined together using dna ligase. Remember: genomic dna is dna that is part of an organism"s chromosomes. A plasmid is small loop of dna which is separate. Cloning vectors: used for isolation and amplification of dna sequences. Mostly originated from bacteria (some from yeast) Replicate independent of cell dna: plasmids can be multi-copy (multiple copies of same gene); i. e. > 1 plasmid/cell. Small, up to about 10 kb (for easy manipulation: maximum insert size: usually ~15 kilo base pairs (kb)

Get access

Grade+
$40 USD/m
Billed monthly
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
10 Verified Answers
Class+
$30 USD/m
Billed monthly
Class+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
7 Verified Answers

Related Documents

MCB 2050 Lecture Notes - Lecture 7: Polymerase Chain Reaction, Southern Blot, Northern Blot
silvergiraffe483
MCB 2050 Lecture Notes - Lecture 15: International Hapmap Project, Avidin, Embryonic Stem Cell
jadeant325
MCB 2050
Final Exam
Study Guide
MCB 2050 Study Guide - Midterm Guide: Polymerase Chain Reaction, Multiple Cloning Site, Agarose Gel Electrophoresis
olivegoat3

Related Questions

Which of the following did Erwin Chargaff observe?

A)

Base composition changes as the organism ages.

B)

Base composition of DNA does not vary from one species toanother.

C)

In all species the number of adenines equals the number ofguanines.

D)

In all species A + T = G + C.

E)

DNAs isolated from different tissues of the same species havethe same base composition.

The melting temperature (Tm) of a DNA duplex is:

A)

the temperature at which double-stranded nucleic acids degradeinto free nucleotides.

B)

the temperature at which double-stranded DNA is toosingle-stranded to ever renature back to double strands.

C)

the temperature at which double-stranded DNA is completelymelted.

D)

the temperature at which double-stranded DNA has become 50%single-stranded.

E)

the temperature at which double-stranded DNA begins to melt.

Given your knowledge and the descriptions of each of thesemethods, which of the following does NOT rely on the ability ofnucleic acids to hybridize with each other?

A)

Southern blotting for the detection of specific DNA fragmentsthat have been separated by gel electrophoresis.

B)

Colony blot hybridization to identify a specific sequence from acollection of sequences that have been inserted into bacteria.

C)

Northern blotting for the detection of specific RNA fragmentsthat have been separated by gel electrophoresis.

D)

A DNA oligonucleotide microarray incubated with probes made fromcDNA.

E)

Western blotting for the detection of specific proteins thathave been separated by gel electrophoresis.

Which of the following is a possible reason why DNA uses thymineinstead of uracil?

A)

Cytidine is deaminated to uracil so it would be difficult forany repair mechanism to differentiate between uracil that belongsin the sequence and uracil that resulted from deamination.

B)

Adenine is deaminated to uracil so it would be difficult for anyrepair mechanism to differentiate between uracil that belongs inthe sequence and uracil that resulted from deamination.

C)

When uracil is bound to a deoxyribose it will become morereactive, forming covalent bonds with other bases.

D)

None of the above.

Which of the following is an enzyme used in cloning to breakcovalent bonds?

A)

DNA ligase

B)

Restriction endonuclease

C)

Reverse transcriptase

D)

DNA polymerase I

E)

Alkaline phosphatase

Which of the following could you use to remove the 5' phosphatesafter cleavage of a plasmid with a single type of restrictionenzyme to ensure that the plasmid would not simply fuse backtogether upon addition of ligase?

A)

Reverse transcriptase

B)

Alkaline phosphatase

C)

DNA ligase

D)

DNA polymerase I

E)

Restriction endonuclease

Which of the following would you use to create a cDNA librarythat you would not use for a genomic library?

A)

Alkaline phosphatase

B)

Reverse transcriptase

C)

Restriction endonuclease

D)

DNA ligase

E)

DNA polymerase I

A plasmid vector typically has which of the followingfeatures?

A)

An origin of replication

B)

A selectable marker to ensure that the only bacteria growingcontain the intact or recombinant vector.

C)

Unique restriction sites for insertion of DNA.

D)

Small size to facilitate entry into the cell and biochemicalmanipulation of the DNA.

E)

All of the above are features of plasmid vectors.

Which of the following is FALSE of genomic or cDNAlibraries?

A)

Genes represented in cDNA libraries include control sequencesthat direct transcription.

B)

cDNA libraries contain cDNAs made from mRNA.

C)

Genomic libraries contain fragments of genomic DNA that aregenerated by partial digests with restriction enzymes.

D)

Genomic libraries usually contain large fragments and are likelyto be constructed in YAC or BAC vectors.

E)

cDNA libraries contain sequences from expressed genes only.

Which of the following is NOT a typical component of apolymerase chain reaction (PCR)?

A)

dNTPs

B)

DNA containing the sequences to be amplified

C)

DNA ligase to connect the fragments together

D)

Primers complementary to each end of the sequence to beamplified

E)

Heat stable Taq polymerase

How can a fusion protein be formed?

A)

Site-directed mutagenesis where a fragment is replaced with asynthetic sequence containing the altered DNA sequence.

B)

Two proteins can recombine in the bacterial cell to form a newhybrid protein.

C)

Portions of two different genes are ligated together to create anew hybrid gene. When expressed the gene product is a fusion of thetwo gene products.

D)

Using oligonucleotide-directed mutagenesis to introduce specificmutations that cause the expressed protein to fuse to otherproteins.

E)

A portion of the gene can be removed to form a shorter gene andtherefore a shorter protein.

Why is green fluorescent protein (GFP) so useful in visualizingfusion proteins in eukaryotes?

A)

The GFP protein recognizes and binds to fusion proteins allowingthe location of the fusion protein to be determined.

B)

The reaction is only dependent on the presence of GFP and oneother cofactor.

C)

It is easily cloned and expressed in bacterial cells.

D)

The presence of just a few molecules of GFP fusion proteins canbe sufficient for observing the proteins microscopically allowingthe study of its location and movements in the cell.

E)

It is derived from fire flies, which are easy to cultivate inthe lab.

You wish to determine the cellular location of a protein butunfortunately it is not particularly antigenic (can t develop astrong antibody response). How could you solve this problem?

A)

Create a fusion protein with an epitope tag that can bevisualized by incubation with fluorescently tagged antibody to theepitope.

B)

Create a fusion with green fluorescent protein (GFP).

C)

Express the protein as a fusion with biotin. The addition of GFPwill cause fluorescence.

D)

A and B might work.

E)

A, B, and C might work.

F)

None of the above will work.

A positive result for the yeast two-hybrid analysis means thefollowing:

A)

The proteins that you are studying can be made into fusionproteins, but do NOT interact with each other within the nucleus ofyeast.

B)

A fusion protein containing the Gal4p activation domain, hasbound to RNA polymerase resulting increase transcription ofnumerous genes, including the reporter gene, and formation ofcolonies.

C)

Two fusion proteins, one containing the Gal4p binding sitedomain and one containing the Gal4p activation domain, haveinteracted through their Gal4p domains resulting in transcriptionof the reporter gene.

D)

Two fusion proteins interact with each other and because of theGal4p binding site domain and the Gal4p activation domain containedin each fusion protein, the reporter genes are activated and thecells die resulting in no colonies.

E)

The colonies growing are expressing the two fusion proteins, butthe two proteins have not interacted with each other.

Which of the following is the approximate size of the humangenome?

A)

3 Gm

B)

300 Kb

C)

3 Mb

D)

3 Gb

E)

3 Mm

Which of the following is correct about the structure orlocation of genes?

Question 22 options:

A)

Prokaryotic genes typically possess their own regulatorysequences that only control expression of a single gene.

B)

Eukaryotic genes often encode introns that are spliced outbefore translation.

C)

Eukaryotic genes are mostly located at the telomeres and thecentromeres of chromosomes.

D)

Eukaryotic genes are typically arranged in operons.

E)

Prokaryotic genes often encode exons that are spliced out beforetranslation